differentiate x(x-3)^2
do I product and chain rule?
f= uv
f'=uv'+vu'
f'= 1*(x-3)^2+ 2x(x-3)*1
shouldn't it be 1 * (x-3)^2 + 2x(x-3)^1 ? instead of 2x(x-3) * 1 at the end?
oh wait nvm i see what i did wrong
x(x-3)^2
=x(x-3)(x-3)
=x(x^2-3x-3x+9)
=x^3-6x^2+9x
=3x^2-12x+9
Yes, to differentiate x(x-3)^2, you would need to use both the product rule and the chain rule. Let me walk you through the steps.
First, let's label the function as f(x) = x(x-3)^2.
To apply the product rule, we'll differentiate each term separately and then combine them.
Step 1: Differentiating the first term "x":
The derivative of x with respect to x is simply 1.
Step 2: Differentiating the second term "(x-3)^2":
Here we need to use the chain rule. Let's call this term g(x) = (x-3)^2.
The chain rule states that if we have a function h(u) and another function g(x) = h(u(x)), then the derivative of g(x) with respect to x is given by:
g'(x) = h'(u) * u'(x)
In this case, h(u) = u^2, and u(x) = x-3.
Differentiating h(u) = u^2 with respect to u gives us h'(u) = 2u.
Differentiating u(x) = x-3 with respect to x gives us u'(x) = 1.
Now, we can use the chain rule:
g'(x) = h'(u) * u'(x)
= 2u * 1
= 2(x-3)
Step 3: Combining the derivatives from steps 1 and 2:
Using the product rule, we multiply the derivative of the first term (1) with the second term unchanged (x-3)^2, and then add the product of the first term (x) with the derivative of the second term (2(x-3)).
f'(x) = (1)(x-3)^2 + (x)(2(x-3))
Now, you can simplify this expression further if desired.